Proving the convergence of a series with very little information

Question

Let $m \in \mathbb{N}$ be a fixed natural number and $ (a_n)_{n \geq 1}$ be a sequence of positive real numbers such that $\forall n \geq 1\colon a_{n+1} \leq a_n - a_{mn}$. Prove that the series $\sum_{n=1}^{\infty}n^\alpha a_n$ converges $\forall \alpha \in \mathbb{R}_+$.

My first idea was to show that $a_n$ is decreasing. From the hypothesis, we have that $ 0 < a_{mn} \leq a_n-a_{n+1}$. So we can sum this equality up from $n=s$ to $n=t$, so we get $\sum_{n=s}^t a_{nm} \leq a_s-a_{t+1}$. However I do not think that it helps that much.

Answer 1

Let $(a_n)$ be any sequence as in OP. We begin by establishing key lemmas.

Lemma 1. For any $l \geq m$, $$ \sum_{n=l}^{\infty} a_n \leq m a_{\lfloor l/m \rfloor}. $$

Proof of Lemma 1. By the positivity of $(a_n)$, we have $a_{n+1} \leq a_n - a_{mn} < a_n$ and hence $(a_n)$ is decreasing. Then for any $l \geq m$, we have

$$ a_{\lfloor l/m \rfloor} \geq \sum_{n=\lfloor l/m \rfloor}^{\infty} (a_n - a_{n+1}) \geq \sum_{n=\lfloor l/m \rfloor}^{\infty} a_{mn} \geq \frac{1}{m} \sum_{n=m\lfloor l/m \rfloor}^{\infty} a_{n}. $$

This proves both assertions of Lemma 1. $\square$

Lemma 2. Define $\xi(\alpha) = \sum_{n=1}^{\infty} n^{\alpha} a_n \in [0, \infty]$. Then

1. $\xi(\alpha) < \infty$ for all $\alpha \leq 0$.
2. For each $\alpha > 0$, there exists a constant $K_{\alpha} \in (0, \infty) $ such that \begin{equation}\label{l2} \xi(\alpha) \leq K_{\alpha} (1 + \xi(\alpha - 1)). \tag{1} \end{equation}

Proof of Lemma 2. Item 1 is immediate from Lemma 2. For Item 2, let $\alpha > 0$. Then there exists a constant $C_{\alpha} \in (0, \infty)$ such that $ n^{\alpha} \leq C_{\alpha} \sum_{k=1}^{n} k^{\alpha-1} $ for all $n \geq 1$. Indeed, this is not hard to prove by noting that $n^{\alpha} = \int_{0}^{n} \alpha t^{\alpha - 1} \, \mathrm{d}t$, and we omit the details. Using this, we get

$$ \xi(\alpha) = \sum_{n=1}^{\infty} n^{\alpha} a_n \leq C_{\alpha} \sum_{n=1}^{\infty} \biggl( \sum_{k=1}^{n} k^{\alpha-1} \biggr) a_n \tag{2} $$

Applying Tonelli's theorem followed by Lemma 1, the sum in the right-hand side of $\text{(2)}$ can be bounded by

\begin{align*} \sum_{n=1}^{\infty} \biggl( \sum_{k=1}^{n} k^{\alpha-1} \biggr) a_n &= \sum_{k=1}^{\infty} k^{\alpha-1} \biggl( \sum_{n=k}^{\infty} a_n \biggr) \\ &= \sum_{k=1}^{m-1} k^{\alpha-1} \biggl( \sum_{n=k}^{\infty} a_n \biggr) + \sum_{k=m}^{\infty} k^{\alpha-1} \biggl( \sum_{n=k}^{\infty} a_n \biggr) \\ &\leq \xi(0) \biggl( \sum_{k=1}^{m-1} k^{\alpha} \biggr) + m \sum_{k=m}^{\infty} k^{\alpha-1} a_{\lfloor k/m \rfloor} \tag{3} \end{align*}

Moreover, it is easy to check that there is a constant $C'_{\alpha} \in (0, \infty)$ satisfying $k^{\alpha} \leq C'_{\alpha}\lfloor k/m \rfloor^{\alpha}$ for all $k \geq m$. Using this, we can further bound $\text{(3)}$ as follows:

\begin{align*} \sum_{n=1}^{\infty} \biggl( \sum_{k=1}^{n} k^{\alpha-1} \biggr) a_n &\leq \xi(0) \biggl( \sum_{k=1}^{m-1} k^{\alpha} \biggr) + m^2 C'_{\alpha-1} \xi(\alpha -1). \tag{4} \end{align*}

Then the desired conclusion follows by combining $\text{(2)–(4)}$. $\square$

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Now the desired assertion easily follows by invoking $\eqref{l2}$ recursively until the argument becomes non-positive.

Answer 2

Given the series $ \sum_{n=1}^{\infty} n^\alpha a_n $ and the condition $ a_{n+1} \leq a_n - a_{mn} $ for all $ n $, we aim to show the series converges for all $ \alpha > 0 $.

First, we establish that $ a_n \to 0 $ as $ n \to \infty $. By summing the inequality $ a_{n+1} \leq a_n - a_{mn} $ from $ n = 1 $ to $ n = k $, we get:

$ \sum_{n=1}^{k} a_{mn} \leq a_1 - a_{k+1} $

Since $ a_n $ is positive and $ a_1 $ is fixed, $ a_{mn} $ must approach zero to prevent the sum from exceeding $ a_1 $.

Next, we apply the Root Test, which considers $ \limsup_{n \to \infty} \sqrt[n]{n^\alpha a_n} $. As $ a_n \to 0 $, the ratio $ \frac{a_n}{n^\alpha} $ and thus the nth root of $ a_n $ approaches zero. Consequently, the nth root of $ n^\alpha a_n $ will also tend to zero for all $ \alpha > 0 $, which implies convergence of the series by the Root Test.

In conclusion, the series $ \sum_{n=1}^{\infty} n^\alpha a_n $ converges for all $ \alpha > 0 $ due to the terms $ a_n $ approaching zero and the application of the Root Test.

Answer 3

This is essentially very similar to Sangchul's solution, but I think the approach might be easier to understand/motivate.

Hints/Observations towards a solution. If you're stuck, explain what you've tried.

1. Show that $a_n$ is a decreasing sequence.
   • OP mentioned this, but did not explicitly show this.
2. For ease of notation (which you'd see in a bit), let $ a_0 = a_1 $ .
   • Note that $n=0$ doesn't satisfy the inequality $ a_{n+1} \leq a_n - a_{nm}$.
3. Show that $ \sum_{i = l}^\infty a_i \leq ma_{\lfloor \frac{l}{m} \rfloor}$.
   • Use OP's observation that $ \sum_{n=s}^t a_{nm} \leq a_s - a_t$.
   • Combine with Observation 1.
   • This is essentially the critical condition. We will iteratively use this to prove the original statement by induction.
4. In particular, for $ \alpha = 0$, $ \sum n^0 a_n \leq ma_0$, so it converges.
   • Yes, I'm aware it's not what the question asked for, but bear with me for a bit.
5. How can we deal with $ \sum n^1 a_n$?
   • Since $n = 1 + 1 + 1 + \ldots 1 $ (n times)...
   • Recognize that we can write this as the double summation $$ \sum_{j=1}^{\infty} \sum _{i=j}^{\infty} a_i \leq \sum_{j=1}^\infty ma_{\lfloor \frac{j}{m} \rfloor} = \sum_{k=1}^\infty m^2 a_k \leq m^3 a_0.$$
6. Show that for $ 0 \leq \alpha \leq 1$, we have $ \sum n^\alpha a_n \leq n^1 a_n \leq m^3 a_0$, so it converges.
7. Can we extend this to $\sum n^2 a_n$?
   • Yes, Use $ n^2 = \sum_{i=1}^{n} 2i-1 \leq \sum_{i=1}^n 2i$
   • What does the double summation look like?
8. Show that the statement is true for $ 0 \leq \alpha \leq 2 $.
9. Can we induct this?
   • Certainly! use $n^k \leq \sum_{i=1}^n kn^{k-1} $ (Sort of a discrete integration, also known as the method of differences)